I got a doubt ....liferocks wrote:x^2+y^2=1 is the equation of circle with radius 1 and center 0...
since a,b is on the circle -1<a<1 and -1<b<1
so b>a+1 or b-a>1 is only possible when the point is in second quadrant
Since the circle x^2+y^2=1 is equally divided in to 4 parts by the X and Y axis the probability that point (a,b) is in second quadrant is [spoiler]1/4[/spoiler]
Ans option A
x^2+y^2=1 is the points on the circle not inside the circle so we don't need the area. You are correct for the point (-1,0) b-a=1 not b-a>1..but second quadrant has infinite number of points so one point not going to make any difference probalility will be 1/4 or almost 1/4.kaulnikhil wrote:I got a doubt ....liferocks wrote:x^2+y^2=1 is the equation of circle with radius 1 and center 0...
since a,b is on the circle -1<a<1 and -1<b<1
so b>a+1 or b-a>1 is only possible when the point is in second quadrant
Since the circle x^2+y^2=1 is equally divided in to 4 parts by the X and Y axis the probability that point (a,b) is in second quadrant is [spoiler]1/4[/spoiler]
Ans option A
Let us consider the equation as b=a+1
now this equations cuts the circle at (-1,0 and 0,1).. Now in case B>a+1
arnt we supposed to find the area of the region enclosed the second quadrant by the line and the circle ?? the area i am referring is above the line b=a+1 ans below the circular region x^2+y^2=1 in second quadrant.. In that case the probablity has to be less than 1/4
could you please clarify?gmatmachoman wrote:(x,y) can be
(1,0)
(0,1)
(-1,0)
(0,-1)
For b>a+1
(-1,1) is the only set that satisfie sthe condition.
p = 1/4.
